Query_12

#$&*

course MTH-279

Professor Smith,I am just sending along some of the work I completed during tutoring, leading up to test two, which I have completed, so I don't lose out on any significant homework credit.

Query 12 Differential Equations

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Question: A cylinder floating vertically in the water has radius 50 cm, height 100 cm and uniform density 700 kg / m^3. It is raised 10 cm from its equilibrium position and released. Set up and solve a differential which describes its position as a function of clock time.

The same cylinder, originally stationary in its equilibrium position, is struck from above, hard enough to cause its top to come just to but not below the level of the water. Solve the differential equation for its motion with this condition, and use a particular solution to determine its velocity just after being struck.

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Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Self-critique rating:

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Question:

For each of the equations and initial conditions below, find the largest t interval in which a solution is known to exist:

• y '' + y ' + 3 t y = tan(t), y(pi) = 1, y ' (pi) = -1

• t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0, y(1) = 0, y ' (1) = 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

• y '' + y ' + 3 t y = tan(t), y(pi) = 1, y ' (pi) = -1

Plotted graph of the tangent function with asymptotes (-3pi/2, -pi/2), (-pi/2, pi/2), (pi/2, 3pi/2). Since we are looking for t = pi, it falls under the third interval (pi/2, 3pi/2)

• t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0, y(1) = 0, y ' (1) = 1

divide by t to get

y’’ + (sin(2t)y’ / t^3 - 9t) + 2y/t = 0

t is undefined at 0

factoring we get, t(t+3)(t-3)

so, t is also undefined at -3 and 3

so we get intervals (neg. infinity, -3), (-3, 0), (0, 3), (3, infinity)

since we are looking at t of 1, this falls under the interval (0, 3)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

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Self-critique rating: OK

Query_12

#$&*

course MTH-279

Professor Smith,I am just sending along some of the work I completed during tutoring, leading up to test two, which I have completed, so I don't lose out on any significant homework credit.

Query 12 Differential Equations

*********************************************

Question: A cylinder floating vertically in the water has radius 50 cm, height 100 cm and uniform density 700 kg / m^3. It is raised 10 cm from its equilibrium position and released. Set up and solve a differential which describes its position as a function of clock time.

The same cylinder, originally stationary in its equilibrium position, is struck from above, hard enough to cause its top to come just to but not below the level of the water. Solve the differential equation for its motion with this condition, and use a particular solution to determine its velocity just after being struck.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Self-critique rating:

*********************************************

Question:

For each of the equations and initial conditions below, find the largest t interval in which a solution is known to exist:

• y '' + y ' + 3 t y = tan(t), y(pi) = 1, y ' (pi) = -1

• t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0, y(1) = 0, y ' (1) = 1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

• y '' + y ' + 3 t y = tan(t), y(pi) = 1, y ' (pi) = -1

Plotted graph of the tangent function with asymptotes (-3pi/2, -pi/2), (-pi/2, pi/2), (pi/2, 3pi/2). Since we are looking for t = pi, it falls under the third interval (pi/2, 3pi/2)

• t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0, y(1) = 0, y ' (1) = 1

divide by t to get

y’’ + (sin(2t)y’ / t^3 - 9t) + 2y/t = 0

t is undefined at 0

factoring we get, t(t+3)(t-3)

so, t is also undefined at -3 and 3

so we get intervals (neg. infinity, -3), (-3, 0), (0, 3), (3, infinity)

since we are looking at t of 1, this falls under the interval (0, 3)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating: OK

#*&!

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